The diprotic acid $H X_{2} A$ has $K_{a 1} = 4.0 \times 1 0^{- 3}$ and $K_{a 2} = 1.0 \times 1 0^{- 9}$ . For a $0.100 M$ solution, neglect the second ionisation and solve the first-step quadratic exactly. Find the pH (two decimals).

Numerical answer type, tolerance 0.01.

Answer: 1.742327

Second ionisation is negligible. Solve

x^{2} / (0.100 - x) = 4.0 \times 1 0^{- 3}

x = \frac{1}{2} [- K + K^{2} + 4 K C] = 0.01810 M

. Hence

pH = - lo g x = 1.74